The Electroweak Theory Based on SU (4) L XU (1) X Gauge Group

Home , Fayyazuddin

arXiv:1605.07835v2 [hep-ph] 6 Jul 2016 lcrwa hoybsdo SU on based theory Electroweak 4 nttt fPyis ita cdm fSineadTechno and Science of Academy Vietnam Physics, of Institute o u hn nvriy oCiMn iy700,Vietnam 700000, City Minh Chi Ho University, Thang Duc Ton 1 hoeia atcePyisadCsooyRsac Group Research Cosmology and Physics Particle Theoretical a a nvriy ue a,SnL 600 Vietnam 360000, La Son Tam, Quyet University, Bac Tay 2 aut fApidSine,TnDcTagUniversity, Thang Duc Ton Sciences, Applied of Faculty u a nvriy aNn iy500,Vietnam 550000, City Nang Da University, Tan Duy .N Long, N. H. 0DoTn aDn,Hni100,Vietnam 100000, Hanoi Dinh, Ba Tan, Dao 10 5 aut fMathematics-Physics-Informatics, of Faculty 3 nttt fRsac n Development, and Research of Institute oCiMn iy700,Vietnam 700000, City Minh Chi Ho ,2, 1, ∗ .T Hue, T. L. 1 ,4, 3, (4) † n .V Loi V. D. and L ⊗ U (1) X ,5, 4, ag group gauge ‡ logy, , Abstract This paper includes two main parts. In the first part, we present generalized gauge models based on the SU(3)C SU(4)L U(1)X (3-4-1) gauge group with arbitrary electric charges of ⊗ ⊗ exotic leptons. The mixing matrix of neutral gauge bosons is analyzed, and the eigenmasses and eigenstates are obtained. The anomaly-free as well as matching conditions are discussed precisely. In the second part, we present a new development of the original 3-4-1 model [2, 9]. Different from previous works, in this paper the neutrinos, with the help of the scalar decuplet H, get the Dirac masses at the tree level. The vacuum expectation value (VEV) of the Higgs boson field in the decuplet H acquiring the VEV responsible for neutrino Dirac mass leads to mixing in separated pairs of singly charged gauge bosons, namely the Standard Model(SM) W boson and K, the new gauge boson acting in the right-handed lepton sector, as well as the singly charged bileptons X and Y . Due to the mixing, there occurs a right-handed current carried by the W boson. From the expression of the electromagnetic coupling constant, ones get the limit of the sine-squared of

2 the Weinberg angle, sin θW < 0.25 and a constraint on electric charges of extra leptons. In the limit of lepton number conservation, the Higgs sector contains all massless Goldstone bosons for massive gauge bosons and the SM-like Higgs boson. Some phenomenology is discussed.

PACS numbers: 12.10.Dm, 12.60.Cn, 12.60.Fr, 12.15.Mm

∗Electronic address: [email protected]

†Electronic address: [email protected]

‡Electronic address: [email protected]

2 I. INTRODUCTION

The current status of particle physics leads to widespread evidence for extending the SM. The recently observed 750 GeV diphoton excess [1] can be explained as an existence of a new neutral scalar that couples to extra heavy quarks, or in some cases, to new leptons and bosons. In this sense, the 3-3-1 models [2–4] seem to be good candidates since they contain all ingredients such as extra quarks and new scalar fields. However, the problem is that satisfying the LHC diphoton excess and the experimental value of the muon anomalous magnetic moment (g 2) requires that the 3-3-1 scale be low ω 400 GeV, while the flavor- − µ ≈ changing neutral current (FCNC) requires a high scale ω 2 TeV. To solve this puzzle for ≈ the 3-3-1 model with right-handed neutrinos, one must introduce an inert scalar triplet [5], extend the gauge group to a larger one such as SU(3) SU(3) SU(3) U(1) [6], or C ⊗ L ⊗ R ⊗ X introduce new charged scalars [7]. The above situation is also correct for the other SM extensions; hence the search for the models satisfying current experimental data is needed. It is known that the SU(4) is the highest symmetry group in the electroweak sector [8]. There have been some gauge models based on the SU(3) SU(4) U(1) [2, 9, 10]; however, the Higgs physics - currently the C ⊗ L ⊗ X most important sector, has not received enough attention. In some versions, this puzzle was not studied in much detail at all. In light of the current status of particle physics, the Higgs sector should be considered with as much detail as possible, especially in the neutral scalar sector where the SM-like Higgs boson is contained. Thus in this work, we will focus on the Higgs sector. As in the 3-3-1 models, the electric charge quantization was also explained in the framework of the 3-4-1 model [11]. The aim of this paper is to present the 3-4-1 models which are able to deal with current Higgs physics. It is well known that if scalar sector contains many neutral scalar fields the situation is very complicated. In addition, we will pay attention to the gauge boson sector where new physics is quite rich and explore some interesting features of phenomenology. The 3-4-1 model we consider here may be considered as combination of the minimal 3-3-1 model [3] and an alternative version with right-handed neutrinos [2, 4]. Hence, the derived model is quite interesting and deserves further study. Our work is arranged as follows. Section II will present conditions of anomaly cancellation and particle content, where extra novel induced leptons have arbitrary electric charges q and

3 q′. We will show that the 3-4-1 models are anomaly free only if there are an equal number of

quadruplets (4) and antiquadruplets (4∗). Simply speaking, this condition requires the sum of all fermion charges to vanish (see subsection II A). In subsection IIB, the Higgs bosons needed for fermion mass production are discussed. We want to avoid nonrenormalizable effective couplings, so every Higgs multiplet has only one component with nonzero VEV. Subsection II Cfocus on the gauge boson fields, especially neutral ones. Section III is devoted to the original 3-4-1 model [2, 9]. As in [9], to produce the mass for leptons, the Higgs decuplet is introduced. However, as will be seen below, it has to be redefined. In this work, we will build a lepton number operator from which the lepton flavor number violating (LFV) processes will be pointed out. In this section, mixing of the singly charged gauge bosons and the currents will get more attention. For completeness, the neutral gauge boson sector will also be presented, though it is quite similar to the previous analysis. From an expression of the ratio of two gauge couplings t, it follows the bound on the sine-squared of the Weinberg 2 angle sin θW < 0.25. In section V the bounds on masses of new gauge bosons are roughly derived, based on the data of W bosons, rare muon decays and µ e conversion. Finally, − we present our conclusion in Sec. VI.

II. THE MODEL

As above mentioned, we ﬁrst check the conditions for anomaly free of the models based on SU(3) SU(4) U(1) gauge group. C ⊗ L ⊗ X

A. Anomaly cancellation and fermion content

For the class of the models constructed from the gauge group SU(3) SU(3) U(1) , C × L × N the conditions for anomaly cancellation were discussed in detail in [12]. Similarly for the class of the SU(3) SU(4) U(1) (3-4-1) models the following gauge anomalies must C × L × X vanish: i) [SU(3) ]2 U(1) , ii) [SU(4) ]3, iii) [SU(4) ]2 U(1) ; iv) [Grav]2 U(1) ; and C × X L L × X × X 3 v) [U(1)X ] . Being diﬀerent from the studies of anomaly cancellation for the 3-3-1 models [12, 13], we will exploit the relation between charge operator and diagonal generators of the gauge symmetry SU(4)L to prove that the ﬁve conditions will reduce to two conditions only: [SU(4) ]3 and [SU(4) ]2 U(1) . L L × X

4 For a general 3-4-1 model, the electric charge operator is in the form

Q = T3 + bT8 + cT15 + X, (1)

where the coeﬃcient in front of T3 equaling 1, is chosen to ensure that the SM group is a subgroup of the model under consideration: SU(2) U(1) SU(4) U(1) . L ⊗ Y ⊂ L ⊗ X The leptons are in quadruplet

q q′ T faL =(νa ,la , Ea , Ea′ )L, a = e, µ, τ , (2) where q and q′ are electric charges of associated extra leptons. Applying Eq.(1) to Eq.(2) we obtain 2q 1 q 3q′ 1 q + q′ 1 b = − − , c = − − , Xf = − , (3) √3 √6 aL 4 or 1 √3b 1 b √2c 1 b c q = , q′ = and Xf = . (4) −2 − 2 −2 − 2√3 − √3 aL −2 − 2√3 − 2√6 Before discussing anomaly cancellation, we remember that the fermion representations of

SU(3)C and SU(4)L in the 3-4-1 models are all SU(3)C triplets, SU(4)L (anti) quadruplets, and singlets. All singlets do not contribute to anomalies so in consideration we omit them.

The representative matrices of generators corresponding to the SU(3)C triplets are denoted a a a as TC (a =1, 2, .., 8) , and the SU(4)L (anti)quadruplets are TL(T L), a =1, 2, .., 15. Now we consider the 3-4-1 model with M and N families of leptons and quarks, respectively. In addition, the number of SU(4)L quadruplets of quark families is K. For simplicity we assume that all left-handed leptons are in the quadruplets. The general case is derived easily. All of them respect the gauge symmetry SU(3) SU(4) U(1) as follows. For C × L × X leptons, we have

′ q q T f = (ν ,l , E , E′ ) (1, 4,X ), iL iL iL iL iL ∼ fL q q′ ν (1, 1, 0), l (1, 1, 1), E (1, 1, q), E′ (1, 1, q′), i =1, 2, ..., M. (5) iR ∼ iR ∼ − iR ∼ iR ∼ These left-handed leptons are generalized from (2). As in the 3-3-1 models [2–4], the parameters b and c are closely connected with q and q′; in [14, 15], the parameters b, c have been used. In our point of view, the clearer way is using q and q′. We note that νiR is an q q′ option, while EiR and EiR may disappear in the speciﬁc case of the minimal 3-4-1 model. If the left-handed leptons are in antiquadruplet, then b and c will be replaced by b and c, − − respectively.

5 The quark sector is

T Q = (u ,d , T , T ′ ) (3, 4,X ), m =1, 2, ..., K, mL mL mL mL mL ∼ qL T Q = (d , u ,D ,D′ ) (3, 4∗,X ), nL nL − nL nL nL ∼ q¯L

upR (3, 1, 2/3), dpR (3, 1, 1/3), TmR (3, 1,XT ), T ′ (3, 1,XT ′ ), ∼ ∼ − ∼ R mR ∼ R

DnR (3, 1,XD ),D′ (3, 1,XD′ ), n = K +1, .., N, p =1, 2, .., N. (6) ∼ R nR ∼ R

3 Let us ﬁrst consider the anomaly of [SU(4)L] . Each of the SU(4)L quadruplets 4L abc a b c or antiquadruplets 4∗ contributes a well-known quantity (4 ) = Tr(T T , T ) or L A L L{ L L} abc a b c (4∗ ) = Tr(T¯ T¯ , T¯ ), where a, b, and c mean three SU(4) gauge bosons related A L L{ L L} L abc abc to the triangle diagrams. Because (4 ) = (4∗ ), the total contribution to the A L −A L 3 [SU(4)L] anomaly can be written as

abc abc (4L) 4L 4∗ = (4L) n n ∗ , (7) A − L A 4L − 4L QmL,fiL QnL ! X X ∗ where n4L and n4L are the number of fermion quadruplets and antiquadruplets, respectively. This means that the above anomaly cancels only if the number of quadruplets is equal to the number of antiquadruplets, namely

M +6K =3N, (8)

where the factor 3 appears because the quarks are in SU(3)C triplets while leptons are in

SU(3)C singlets. Next, we consider the anomaly of [SU(4) ]2 U(1) . The [SU(4) ]2 gives the same L × X L a b a b factor for both quadruplets and anti - quadruplets, i.e., Tr[T T ] = Tr[T¯ T¯ ]= δab/2, where a and b relate to two SU(4)L gauge bosons. Hence, this anomaly is free if the sum of all

U(1)X charges of the SU(4)L chiral multiplets is zero, namely,

X = MX +3KX + 3(N K)X =0, (9) L fL qL − q¯L f ,Q iLXpL where XL denotes the U(1)X charge of an arbitrary left-handed (anti) - quadruplet in the model. Now we turn to the anomaly of the [SU(3) ]2 U(1) . This case is similar to the case C × X of [SU(4) ]2 U(1) , but now only the SU(3) quark triplets contribute to the mentioned L × X C

6 anomaly. The anomaly-free condition is

4X + 4X X =0, (10) qL q¯L − qR Q Q q XmL XnL XR

where XqR and XqL,q¯L are U(1)X charges of right-handed SU(4)L quark singlets qR and left-

handed SU(4)L quark (anti)quadruplets QmL (QnL ), respectively. The factors 4 appear in

Eq. (10) because we take four components of every SU(4)L (anti)quadruplet into account. The minus sign implies the opposite contributions of left- and right-handed fermions to gauge anomalies. Because all q are singlets of the SU(4) U(1) , their U(1) charges R L × X X are always equal to the electric charges qqR , leading to qR XqR = qR qqR . On the other hand, from the deﬁnition of the charge operators Q givenP in (1), it canP be seen that

4X = Tr (X I )= Tr[Q], qL L × 4 Q Q Q XmL XmL XmL where Tr[Q] implies the sum over the electric charges of all components of the quark quadruplet QmL. Note that we have used the traceless property of the SU(4)L generators: Tr(T a) = 0. Doing this the same way for the case of quark antiquadruplets, the condition (10) can be rewritten in terms of the electric charges of left-and right-handed quarks,

4 q q =0, (11) qL − qR Q i=1 q XpL X XR where the ﬁrst sum implies that all SU(4)L quark (anti)quadruplets and their components are counted. The equality (11) is always correct because every left-handed quark always has its right-handed partner with the same electric charge. The above discussion on the anomaly cancellation of [SU(3) ]2 U(1) can be applied C × X for the case of the anomaly cancellation of [Grav]2 U(1) , but the electric charges must be × X counted for all components of the SU(3)C and SU(4)L multiplets of all quarks and leptons. The proof of the zero contribution of the quark sector is very easy while that of the lepton sector needs more explanation. Although the presence of right-handed neutrinos is optional, they are neutral leptons and, therefore, do not contribute to this anomaly. If right-handed charged leptons are arranged into components of left-handed (anti)quadruplets, they must be changed into their charge conjugations. These new forms of right-handed leptons have the opposite signs of electric charges compared to the respective left-handed partners. Hence, the total contribution to the considered anomaly is still zero.

7 3 Cancellation of the [U(1)X ] anomaly, which relates to the triangle diagram having three

B′′ gauge bosons, is expressed by the following condition:

X3 X3 =0, (12) FL − FR F F XL XR where FL and FR are any components of the fermion (quark and lepton) representations of

the SU(3)C and SU(4)L gauge symmetries. Hence, the sum is taken over all components

of these representations. Because the U(1)X and the electric charges relate to each other through the deﬁnition of the charge operator (1), we can write the left-hand side of (12) as a function of electric charges. Because all right-handed fermions are SU(4)L singlets,

XFR = qFR ; therefore, 3 3 XFR = qFR . (13) F F XR XR In contrast, all left-handed fermions are (anti)quadruplets, and we can write the left-handed term in (12) as a sum over all fermion (anti)quadruplets, namely,

3 3 XFL = 4XFL . (14) F , ∗ XL 4XL 4L

Now we come back to the formula of the charge operator, where we denote QFL as the charge operator of the left-handed fermion (anti)quadruplets. If we denote T (3,8,15) T 3 + bT 8 + ≡ 15 cT , the U(1)X charge part of each 4L representation can be written as

3 X I = Q T (3,8,15) (X I )3 = Q T (3,8,15) FL 4 FL − → FL 4 FL − 3 Tr X3 I = Tr Q3 3Tr Q T (3,8,15)(Q T (3,8,15)) Tr T (3,8,15) → FL 4 FL − FL FL − − h i 3 4X3 = Tr Q3 3Tr (T (3,8,15) + X I )T (3,8,15)X I Tr T (3,8,15) → FL FL − FL 4 FL 4 − 3 h i = Tr Q3 3X Tr (T (3,8,15))2 Tr T (3,8,15) , (15) FL − FL − h i where we have used the fact that both Q and T (3,8,15)are diagonal so they commute with each (3,8,15) 3 other, and T is traceless. Remember that 4XFL is the contribution of four components 3 in one quadruplet. Then the contribution to [U(1)X ] of all quadruplets is

3 X3 = q3 3 X (T (3,8,15))2 n Tr T (3,8,15) . (16) FL FL − FL − 4L 4 F 4 XL XL XL h i a The same proof can be applied for the case of antiquadruplets with generators T = T a, − (3,8,15) (a =3, 8, 15) and T = T (3,8,15). From this, it can be proved that − (3,8,15) 2 2 (3,8,15) 3 3 Tr T = Tr T (3,8,15) , Tr T = Tr T (3,8,15) . − h i h i 8 The above discussion is enough to write (12) in the following new form:

3 3 3 (3,8,15) 2 (3,8,15) ∗ qFL qFR 3Tr (T ) XFL Tr T n4L n4L =0. (17) − − ∗ − − FL FR ! 4L,4 X X XL h i The equality (17) is satisﬁed as a consequence of the two anomaly-free conditions (7) and (9). The equality (9) also implies that the sum over the electric charges of left-handed fermions is zero. Finally, we conclude that the anomaly-free conditions of the 3-4-1 models are as follows: (i) the number of fermion quadruplets is equal to that of fermion antiquadruplets, (ii) sum over electric charges of all left-handed fermions is zero. The 3-4-1 models, of concern here are all satisﬁed with these two conditions.

B. Yukawa couplings and masses for fermions

Since the leptons are arranged as

q q′ T 1 f = (ν ,l , E , E′ ) 1, 4, (q + q′ 1) , aL a a a a L ∼ 4 − q q′ l (1, 1, 1) , E (1, 1, q) , E′ (1, 1, q′) , (18) aR ∼ − aR ∼ aR ∼

q′ the mass of Ea′ is obtained from the Yukawa coupling,

E′ E′ q′ L = h f Φ E′ + H.c., (19) − Yukawa ab aL 1 bR

where ′ ′ ′ T (q 3q′ 1) ( q ) ( q 1) (q q ) 0 Φ 1, 4, − − = Φ − , Φ − − , Φ − , Φ . (20) 1 ∼ 4 1 1 1 1 ′ Hence, if Φ0 has a VEV V , then E q gets mass from a mass matrix 1 √2 a′

E′ V (mE′ )ab = h . (21) ab √2

q The mass of Ea is obtained from the following Yukawa term,

LE = hE f Φ Eq + H.c., (22) − Yukawa ab aL 2 bR

where ′ T (1+3q q′) ( q) ( q 1) 0 (q q) Φ 1, 4, − = Φ − , Φ − − , Φ , Φ − . (23) 2 ∼ − 4 2 2 2 2 9 Thus, if Φ0 has a VEV ω , then Eq gets mass from a matrix: 2 √2 a

E ω (mE)ab = h . (24) ab √2 Finally the ordinary lepton masses come from the following Yukawa term,

Ll = hl f Φ l + H.c., (25) − Yukawa ab aL 3 bR where (3 + q + q ) ′ T Φ 1, 4, ′ = Φ(+) , Φ0 , Φ(q+1) , Φ(q +1) . (26) 3 ∼ 4 3 3 3 3 If Φ0 has a VEV v , then the mass matrix related to masses of l is 3 √2 a

l v (ml)ab = h . (27) ab √2 We turn now to the quark sector where

T 5+3(q + q′) Q = (u ,d ,T,T ′) 3, 4, , 3L 3 3 L ∼ 12 u (3, 1, 2/3) , d (3, 1, 1/3) , 3R ∼ 3R ∼ − 2+3q 2+3q′ T 3, 1, , T ′ 3, 1, . (28) R ∼ 3 R ∼ 3

The u3 gets mass through the Yukawa part,

Lt = htQ Φ u + H.c., (29) − Yukawa 3L 4 3R

where ′ T (q + q′ 1) 0 (q) (q ) Φ 1, 4, − = Φ , Φ− , Φ , Φ . (30) 4 ∼ 4 4 4 4 4 If Φ0 has a VEV u , then the mass term of u is 4 √2 3

t u mu3 = h . (31) √2

The other Yukawa terms related to Q3L are

g3 b T T ′ L = h Q Φ d + h Q Φ T + h Q Φ T ′ + H.c., (32) − Yukawa 3L 3 3R 3L 2 R 3L 1 R which give three mass terms:

v ω ′ V b T ′ T md3 = h , mT = h , mT = h . (33) √2 √2 √2

10 Two other quark generations are

T 1+3(q + q′) Q = (d , u ,D ,D′ ) 3, 4∗, , α =1, 2, αL α − α α α L ∼ − 12 u (3, 1, 2/3) , d (3, 1, 1/3) , αR ∼ αR ∼ − 1+3q 1+3q′ D 3, 1, ,D′ 3, 1, . (34) αR ∼ − 3 αR ∼ − 3 The relevant Yukawa terms are

12 d2 u2 L = h Q Φ† d + h Q Φ† u + − Yukawa αβ αL 4 βR αβ αL 3 βR D2 D′2 + hαβ QαLΦ2† DβR + hαβ QαLΦ1† DβR′ + H.c., (35) from which it follows that

u v ω ′ V d2 u2 D2 ′ D 2 (md2 )αβ = h , (mu2 )αβ = h , (mD2 )αβ = h , (mD )αβ = h . (36) αβ √2 − αβ √2 αβ √2 2 αβ √2 We emphasize that if all fermions except neutrinos have the right-handed counterparts, then only four Higgs quadruplets are needed. Because the sum of the X-charges over four Higgs quadruplets vanishes, in the Higgs potential, there always exists an antisymmetric term i j k l ǫijklΦ1Φ2Φ3Φ4.

C. Gauge boson masses

Gauge boson masses arise from the covariant kinetic term of the Higgs bosons,

4 µ L = (D Φ )† D Φ . (37) Higgs h ii µh ii i=1 X The covariant derivative is deﬁned as 15 D = ∂ ig A T ig′XB′′T µ µ − aµ a − µ 16 a=1 X ∂ igP NC igP CC , (38) ≡ µ − µ − µ

where g,g′ and Aaµ, Bµ′′ are gauge couplings and ﬁelds of the gauge groups SU(4)L and U(1) , respectively. For the quadruplet, T = 1 diag(1, 1, 1, 1), and the part relating to X 16 2√2 neutral currents is 1 A A B A A B P NC = diag A + 8 + 15 + Xt ′′ , A + 8 + 15 + Xt ′′ , µ 2 3 √ √ √ − 3 √ √ √ 3 6 2 3 6 2 2A A B 3A B 8 + 15 + Xt ′′ , 15 + Xt ′′ , (39) − √ √ √ − √ √ 3 6 2 6 2µ 11 where the spacetime indices of gauge ﬁelds are omitted for compactness, and t g′/g. The ≡ part associated with charged currents is 1 P CC = λ A ; a =1, 2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14 µ 2 a aµ a X ′ + q q 0 W W13− W14− (1+q) (1+q′) 1  W − 0 W − W −  = 23 24 , (40) q (1+q) (q q′) √2  W W 0 W −   13 23 34   q′ (1+q′) (q q′)   W W W − − 0   14 24 34 µ   where we have denoted √2 W µ Aµ iAµ and so forth. The upper subscripts label the 13 ≡ 1 − 3 electric charges of gauge bosons. We note that this part does not depend on the X-charges of quadruplets. To summarize, with the following Higgs vacuum structure, V T ω T Φ = 0 , 0 , 0 , , Φ = 0 , 0 , , 0 , h 1i √ h 2i √ 2 2 v T u T Φ = 0 , , 0 , 0 , Φ = , 0 , 0 , 0 , (41) h 3i √ h 4i √ 2 2 masses of non-Hermitian (charged) gauge bosons are given by g2(v2 + u2) g2(u2 + ω2) g2(v2 + ω2) m2 = , m2 = , m2 = , W 4 W13 4 W23 4 g2(u2 + V 2) g2(v2 + V 2) g2(ω2 + V 2) m2 = , m2 = , m2 = . (42) W14 4 W24 4 W34 4 By spontaneous symmetry breaking (SSB), the following relation should be in order: V ≫ ω u, v, and from (42) one gets ≫ 2 2 2 2 2 u + v = vSM = 246 GeV . (43)

D. Neutral gauge bosons

Inserting Eq.(39) into the Higgs multiplets, we get mass terms g2 A A B 2 M 2NG = u2 A + 8 + 15 + X t ′′ mass 4 3 √ √ Φ4 √ " 3 6 2 A A B 2 + v2 A + 8 + 15 + X t ′′ − 3 √ √ Φ3 √ 3 6 2 2A A B 2 3A B 2 + ω2 8 + 15 + X t ′′ + V 2 15 + X t ′′ . (44) − √ √ Φ2 √ − √ Φ1 √ 3 6 2 6 2 #

12 In the basis A3µ, A8µ, A15µ, Bµ′′ , the respective squared mass matrix is given by g2 M 2NG = mass 4 u2 + v2 1 (u2 v2) 1 (u2 v2) t (X u2 X v2) √3 − √6 − √2 Φ4 − Φ3   1 2 2 2 1 2 2 2 t 2 2 2 3 (u + v +4ω ) √ (u + v 2ω ) √ (XΦ4 u + XΦ3 v 2XΦ2 ω )  3 2 − 6 −  .    1 (u2 + v2 + ω2 +9V 2) t (X u2 + X v2 + X ω2 3X V 2)   6 2√3 Φ4 Φ3 Φ2 − Φ1   2   t 2 2 2 2 2 2 2 2   (XΦ4 u + XΦ3 v + XΦ2 ω + XΦ1 V )   2   (45)

Following the above assumption, the SSB following the pattern

V ω u,v SU(4) U(1) SU(3) U(1) SU(2) U(1) U(1) L ⊗ X −→ L ⊗ N −→ L ⊗ Y −→ Q will be used for constructing the matching relation of the gauge couplings and U(1) charges of the group SU(4) U(1) and those of the SM gauge group SU(2) U(1) . Corresponding L × X L × Y to each step of the breaking, the neutral gauge boson states will be changed as follows:

′′ A3,A8,A15,B SU(4)L U(1)X SU(3)L U(1)N ′ ⊗′′ −→ ′ ′ ⊗ A3,A8,B ,Z4 A3,B,Z3,Z4 SU(2) U(1) U(1) : A,Z,Z′ ,Z′ . (46) −→ L ⊗ Y −→ Q 3 4 At the ﬁrst step of breaking, the nonzero VEV V = 0 just results in M 2NG M 2 = 6 mass → 43 M 2NG , where the M 2 is mass |w=v=u=0 43 00 0 0

2 2 g  00 0 0  M43 = 2 2 . (47) 3V 3ctV 4  0 0 −   2 4√2   3ctV 2 3c2t2V 2   0 0 −   4√2 16    The transformation C43 relating the two before- and after-breaking bases, namely T T T (A , A , A , B′′) = C (A , A , B′,Z′′) , is given by 3 8 15 43 × 3 8 4 10 0 0  01 0 0  C43 = (48)  0 0 c s   43 43     0 0 s c   − 43 43    with ct 2√2 c43 and s43 = . (49) ≡ √8+ c2t2 √8+ c2t2

13 After this step, only A15 and B′′ mix with each other to create the U(1)N gauge boson, denoted as B′, of the SU(3) U(1) group. Also, the case c = 0 leads to c =0 and L × N 43 s43 = 1, implying that the SU(4)L decouples from the U(1)X . The diagonal squared mass 2 2 2 T 3V matrix can be found as M = C43M C = diagonal 0, 0, 0, 2 , including three massless 43d 43 43 2s43 and one massive values. Similarly, the second step of the breaking from SU(3) U(1) SU(2) U(1) can be L × N → L × Y T T T done by the second transformation C satisfying (A , A , B′,Z′′) = C (A ,B,Z′′,Z′′) , 32 3 8 4 32 × 3 3 4 where Z3′′ and Z4′′, at this moment, are not mass eigenstates. The squared mass matrix 2NG 2 2NG T T T now is M M = M so that (A , A , A , B′′) = C (A ,B,Z′′,Z′′) . mass → 42 mass |v=u=0 3 8 15 42 × 3 3 4 The concrete transformation of the two steps of the breaking are C42 = C32.C43. These transformations are given as follows,

1 0 00 10 0 0

 0 c32 s32 0   0 c32 c43s32 s43s32  C32 = , C42 = , (50)  0 s c 0   0 s c c s c   32 32   32 43 32 43 32   −   −   0 0 01   0 0 s c     − 43 43      where

2 2 2√2 √8+ c t bts43 bt s32 = = , c32 = = . (51) 2 2 2 2 2 2 2 2 2 2 2 2 b t s43 +8 8+(b + c )t b t s43 +8 8+(b + c )t After twop steps of breaking,p the squared mass matrixp p

00 0 0

2   2 2 T g 00 0 0 M ′ = C42M C = 2 2 (52) 42 42 42 4w √2( 1+c43s43bt)w 4  0 0 2 −   3s32 3s43s32   2 2 2 2  √2( 1+c43s43bt)w ( 1+c43s43bt) w +9V  0 0 − − 2   3s43s32 6s   43    contains two massless eigenvalues corresponding to the A3 and B gauge bosons of the SM gauge group SU(2) U(1) . The two states Z′′ and Z′′ are still not eigenstates, but they L × Y 3 4 decouple from the SM gauge bosons. They can be easily diagonalized, but that will be done later. To ﬁnd the matching condition, we pay attention to the Bµ state involved with the neutral part of the covariant derivative that changes following the the steps of breaking,

14 namely

(41) 3 8 15 16 D = ∂ ig A T + A T + A T + t X B′′T neutralµ µ − 3µ 8µ 15µ × × µ D(21) = ∂ ig A T 3 + c B T 8 + c s B T 15 + t X s s B T 16 → neutralµ µ − 3µ 32 µ 43 32 µ × × 43 32 µ 3 gt 8 15 = ∂µ igA 3µT i Bµ bT + cT + XI4 , (53) − − 8+(b2 + c2)t2 p (21) where all A8,15 and B′′ are replaced by B based on (50). Identifying the Dµ in (53) with

the covariant derivative deﬁned in the SM, we derive that the gauge coupling of the SU(4)L

is the SU(2)L coupling g. The other two important equalities are

gt Y 8 15 = g1, and = bT + cT + XI4, (54) 8+(b2 + c2)t2 2 b p where g1 and Y are the coupling and charge operator of the SM U(1)Y gauge group. The second formula in (54) is consistent with the identification of Y from the definition of the b electric charge operator (1). Furthermore, it can be seen that N cT 15 + X and b b ≡ ≡ β/√3 are relations between parameters defined in the gauge groups SU(4)L U(1)X and b × SU(3) U(1) . L × N 2 From g1/g = sW /cW , where sW =0.231, we find

g 2√2s t = ′ = W . (55) g 1 (1 + b2 + c2)s2 − W We emphasize the ratio of two couplingsp - the parameter t is symmetric in changing from q to q′. To see this, with the help of Eq. (3), the above formula can be written in terms of q and q′ as follows: 2√2s t = W . (56) 1 q + q qq + 3 (1 + q2 + q′2) s2 − ′ − ′ 2 W Keeping in mind that s2 q0.23, from (56) we get a constraint on electric charges of the W ≃ q q′ new exotic leptons Ea and Ea′ , namely

3 2 ′2 q + q′ qq′ + (1 + q + q ) 4 . (57) − 2 ≤

The squared mass matrix (52) can be diagonalized by a matrix C32′ , which gives two mass

eigenstates Z′ and Z′ in the second step of the breaking. While the breaking from SU(2) 3 4 L × U(1)Y to U(1)Q is taken by the well-known transformation C21, the two transformative

15 matrices are

sW cW 0 0 10 0 0

 cW sW 0 0   01 0 0  C21 = − , C32′ = , (58)  0 0 10   0 0 c s     α α       0 0 01   0 0 s c     − α α      where cα = cos α and sα = sin α satisfy

2√2( 1+ c s bt) w2 4√2s s ( 1+ c s bt) w2 t 43 43 = 43 32 43 43 . (59) 2α − 2 2 2 2 2 2 2 − 2 2 2 ≡ − ( 1+c43s43bt) w +9V 4w 8s w s [( 1+ c43s43bt) w +9V ] s43s32 − 2 2 43 32 2s43 − s32 − − h i 2 2 T 2 2 Then we have M = C′ M C′ = diag(0, 0, m ′ , m ′ ), where 42d 32 42 32 Z3 Z4

2 2 2 2 √ 2 2 2 g 3sαV 2 2cαs43 + sαs32 ( 1+ bs43c43t) w m ′ = M = + − , Z3 42d 33 2 2 2 4 " 2s43 6s43s32 # 2 2 2 2 √ 2 2 2 g 3cαV 2 2sαs43 cαs32 ( 1+ bs43c43t) w m ′ = M = + − − . (60) Z4 42d 44 2 2 2 4 " 2s43 6s43s32 # The total transformation after all steps of breaking is C = C21.C32′ .C32.C43. The squared mass matrix of the neutral gauge boson transformed under this rotation is derived as

2 2NG T 2 2 2 ′ ′ M41 = C.Mmass .C = diag(0, 0, mZ3 , mZ4 )+ δM41, (61) where δM 2 is a 4 4 matrix having the property that (δM 2 ) = (m2 ) with all 41 × 41 ij O W 2 2 i, j = 1, 2, 3, 4. In addition, (δM41)i0 = (δM41)0i = 0 with any i = 1, 2, 3, 4, and 2 2 2 (δM41)22 = mZ . We can approximately consider M41 as the diagonal matrix, where C is the transformation relating the original and physical bases of neutral gauge bosons T T (A3, A8, A15, B′′) and (A,Z,Z3,Z4) ; precisely

Aµ = sW A3µ + cW c32A8µ + c43s32A15µ + s43s32Bµ′′ ,

Z c A s c A + c s A + s s B′′ , µ ≃ W 3µ − W 32 8µ 43 32 15µ 43 32 µ Z Z′ = s c A +(c c c s s ) A +(s c c + c s ) B′′, 3µ ≃ 3µ − 32 α 8µ 43 32 α − 43 α 15µ 43 32 α 43 α µ Z Z′ = s s A (c c s + s c ) A +(c c s c s ) B′′. (62) 4µ ≃ 4µ 32 α 8µ − 43 32 α 43 α 15µ 43 α − 43 32 α µ

Now we return to the Higgs content of the model. From the above presentation we explicitly see that:

16 1. If q, q′ =0 , q,q′ = 1, and q = q′, then the Higgs sector is smallest containing only 6 6 − 6 four neutral Higgs ﬁelds.

2. The case q = q′ = 0 has been considered in [16], where the Higgs sector contains ten neutral Higgs ﬁelds, and there are three non-Hermitian neutral gauge. This case is extremely complicated.

3. The case q = q′ = 1 has been considered in [14–20]. −

4. The case q =0, q′ = 1 has been considered in [2, 9, 21].

5. The case q = 1, q′ = 0 has been considered in [21]. − 6. SU(4)(L) x U(1)(X) models with a little Higgs have been presented in [22].

Let us summarize the SSB pattern. At the ﬁrst step of symmetry breaking through V , only the following ﬁelds get masses: the prime fermions including the exotic leptons Ei , quarks

T and quarks Dα; and the gauge bosons W34 and Z. At the second step of SSB through ω, all remain exotic fermions - exotic leptons and quarks get masses. The charged gauge bosons in the top right corner of the non-Hermitian gauge boson matrix [see, (40)] and the

extra Z′ obtain masses. Finally, the last step is possible through u and v and all the SM fermions and gauge bosons get masses. Let us take time to brieﬂy review the development of the SU(4) U(1) models. To our ⊗ knowledge, the ﬁrst attempt by Fayyazuddin and Riazuddin [23] introduced the decuplet.

With the electric charges of leptons q = 0, q′ = 1, the limit on the sine-squared of the 2 Weinberg angle was obtained: sin θW = 0.25 and the bound for uniﬁcation mass is as follows: 3.3 104 m 6.4 103 GeV. At that time, the particle arrangement in × ≥ X ≥ × Ref. [23] was not correct. The next step belongs to M. B. Voloshin [8] who attempted to solve problem related to the realizability of the small mass and large magnetic moment of neutrinos. For this purpose, the author focused on the lepton sector only where the particle arrangement is the same as ours (see below). The 3-4-1 model in the form discussed here was proposed in [2, 9]. The questions con- cerning anomaly cancellation and quantization of electric charge, and the neutrino and generation nonuniversality were addressed in [13] and [24], respectively. In [25], the neutrinos and electromagnetic gauge invariance were discussed, while the Majoron emitting

17 neutrinoless double beta decay in the minimal 3-4-1 model with right-handed neutrino con-

taining a decuplet were addressed in [25, 26]. In association with discrete Z2 symmetry, the model without exotic electric charges providing a consistent mass spectrum, was proposed in [27]. The SU(4)(EW ) U(1)(B l) model with left-right symmetry was proposed in [28]. × − It is interesting that the electroweak uniﬁcation of quarks and leptons in a gauge group SU(3) SU(4) U(1) was built in [29]. The muon anomalous magnetic moment in the C × × SU(4) U(1) model was considered in [30]. The neutrino mass and mixing in the special × N formalism were presented in [10]. It is worth mentioning that, except for the suspersymmetric 3-4-1 model [31], the Higgs potential containing a decuplet is presented for the ﬁrst time in this paper.

Now we turn to the model similar to the one originally built in [2, 9] with q =1, q′ = 0.

III. MINIMAL 3-4-1 WITH RIGHT-HANDED NEUTRINOS

Let us consider a model in which leptons are arranged as

f =(ν ,l ,lc , νc)T (1, 4, 0) , a = e, µ, τ , (63) aL a a a a L ∼ where we have in mind that νc (ν )c and the charge conjugation of f : f c (f )c = L ≡ R aL aR ≡ aL c c T (νaR ,laR, laR, νaR) . One quark generation is in quadruplet:

T 2 Q = (u ,d ,T,T ′) 3, 4, , 3L 3 3 L ∼ 3 u (3, 1, 2/3) , d (3, 1, 1/3) , 3R ∼ 3R ∼ − 5 2 T 3, 1, , T ′ 3, 1, . (64) R ∼ 3 R ∼ 3

5 ′ 2 The exotic quarks have electric charges: qT = 3 , qT = 3 . Two other quark generations are in antiquadruplet

T 1 Q = (d , u ,D ,D′ ) 3, 4∗, , α =1, 2, αL α − α α α L ∼ −3 u (3, 1, 2/3) , d (3, 1, 1/3) , αR ∼ αR ∼ − 4 1 D 3, 1, ,D′ 3, 1, . (65) αR ∼ −3 αR ∼ −3 4 1 The exotic quarks have electric charges: q = , q ′ = . Dα − 3 Dα − 3

18 Applying Eq.(1) to Eq.(63), we obtain

b = √3 , c =0 , X =0. (66) − faL

Then the electric charge operator, for the quadruplet, has the form

Q = diag (X, 1+ X, 1+ X,X) . (67) −

For SSB, we need four Higgs quadruplets, namely,

0 + 0 T 0 T χ = χ , χ− , χ , χ (1, 4, 0) , φ = φ− ,φ−− ,φ ,φ− (1, 4, 1) , 1 2 3 4 ∼ 1 2 ∼ − + 0 ++ + T 0 + 0 T ρ = ρ , ρ , ρ , ρ (1, 4, 1) , η = η , η− , η , η (1, 4, 0) . (68) 1 2 ∼ 1 2 3 4 ∼ In [19], the Higgs sector contains only three Higgs quadruplets, but to produce masses of charged leptons and neutrinos, the nonrenormalizable eﬀective dimension-ﬁve and -nine operators were used. Here we prefer the original way in [2, 9]. The Yukawa couplings for the quark sector are

q t b T T ′ L = h Q ηu + h Q ρd + h Q φ T + h Q χ T ′ − Yukawa 3L 3R 3L 3R 3L R 3L R d2 u2 D2 D′2 + hαβQαLη†dβR + hαβQαLρ†uβR + hαβ QαLφ†DβR + hαβ QαLχ†DβR′ + H.c.. (69)

If the Higgs sector has VEV structure as

V T ω T χ = 0 , 0 , 0 , , φ = 0 , 0 , , 0 , h i √ h i √ 2 2 v T u T ρ = 0 , , 0 , 0 , η = , 0 , 0 , 0 , (70) h i √ h i √ 2 2 then the quarks get masses as follows:

u v ω ′ V t b T ′ T mu3 = h , md3 = h , mT = h , mT = h , √2 √2 √2 √2 u v ω ′ V d2 u2 D2 ′ D 2 (md2 )αβ = h , (mu2 )αβ = h , (mD2 )αβ = h , (mD )αβ = h . (71) αβ √2 − αβ √2 αβ √2 2 αβ √2 Until now, the leptons were massless. To produce masses for leptons from the renormalizable c Yukawa interactions, we will base it on the product f f 6 10∗ . If an antisymmetric aL bR ∼ A ⊕ S 6 is used, then the lepton mass matrix will be antisymmetric; consequently one lepton is

19 still massless. So, the better way is to introduce a symmetric decuplet (10S) given by

0 + 0 H1 H1− H2 H2 0  H1− H1−− H3 H3−  H′ (1, 10, 0) = . (72) ∼  H+ H0 H++ H+   2 3 2 4   0 + 0   H H− H H   2 3 4 4  The gauge-invariant Lagrangian of the 10-plet is given by 

H′ µ L = Tr (D H′)†D H′ V. (73) 0 µ − We will show that the Higgs content written in (72) should be redeﬁned. To clarify this, let

us consider the kinetic part of H′ in (73)

H′ µ Lkinetic = Tr (∂µH′)†∂ H′

0 µ 0 0 µ 0 ++ µ ++ µ = ∂µH1∗∂ H1 + ∂µH4∗∂ H4 + ∂µH1 ∂ H1−− + ∂µH2 ∂ H2−−

0 µ 0 0 µ 0 +2(∂µH2∗∂ H2 + ∂µH3∗∂ H3

+ µ + µ + µ + µ + ∂µH1 ∂ H1− + ∂µH2 ∂ H2− + ∂µH3 ∂ H3− + ∂µH4 ∂ H4−) . (74)

The factor 2 in the second line of (74) shows that nondiagonal ﬁelds in (73) must be redeﬁned as follows: 0 + 0 √2H1 H1− H2 H2 0 1  H1− √2H1−− H3 H3−  H′ H = . (75) → √2  H+ H0 √2H++ H+   2 3 2 4   0 + 0   H H− H √2H   2 3 4 4  If so, we have  

0 0 0 0 0 0 0 0 Tr (H)†H = H1∗H1 + H4∗H4 + H2∗H2 + H3∗H3

++ ++ + + + + +H1 H1−− + H2 H2−− + H1 H1− + H2 H2− + H3 H3− + H4 H4−. (76)

In what follows, we will use H only. The Yukawa interaction for the lepton is given by

l l c LYukawa = habfaLHfbR + H.c. − l hab c 0 c + 0 = νaL √2ν H + l H− + lbRH + νbRH √2 bR 1 bR 1 2 2 h c c 0 + laL νbRH1− + √2lbRH1−− + lbRH3 + νbRH3−

c c + c 0 ++ + + laL νbRH2 + lbRH3 + √2lbRH2 + νbRH4 c c 0 c + √ 0 + νaL νbRH2 + lbRH3− + lbRH4 + 2νbRH4 + H.c.. (77) i 20 As usual, assuming an expansion of the neutral Higgs ﬁelds as follows,

v′ + RH0 iIH0 ǫ + RH0 iIH0 H0 = 3 − 3 , H0 = 2 − 2 , (78) 3 √2 2 √2 then 00 0 ǫ

1  0 0 v′ 0  H = . (79) h i 2  0 v 0 0   ′     ǫ 0 00    The charged leptons get mass matrix given by 

l l hab 0 hab v′ (ml)ab = H = . (80) √2h 3 i 2 The neutrinos obtain the Dirac mass by H0 in the same Yukawa coupling matrix: h 2 i l l hab 0 hab ǫ (mν)ab = H = . (81) √2h 2 i 2 The neutrino Majorana mass will follow from H0 and H0 . h 1 i h 4 i Noting that the mixing parameters among the charged leptons are tiny, while those among the neutrinos are large, the matrix in (81) must be modiﬁed. It is hoped that the radiative corrections will provide the mixing that matches the current experimental data [32]. We will return to this problem in our future work. It is emphasized that there are ﬂavor-lepton violating interactions in Eq. (77). The lepton number operator is constructed from the diagonal generators as follows,

L = αT + βT + γT + . (82) 3 8 15 L

For the general case, let us assume that the new extra leptons E and E′ acquire lepton

number l and l′, respectively,

T faL =(νa ,la , Ea , Ea′ )L, a = e, µ, τ . (83)

Applying (82) for (83), we obtain

1 1 2(1 l) (2 + l 3l′) α =0 , f = + (l + l′) , β = − ,γ = − . (84) L aL 2 4 √3 √6 The vanishing of the coeﬃcient α is a consequence of the lepton number conservation in the SM. Thus, 2(1 l) (2 + l 3l′) L = − T8 + − T15 + . (85) √3 √6 L

21 TABLE I: and charges for the multiplets in the 3-4-1 model with right-handed neutrinos. B L

Multiplet χφη ρH Q3L QαL uaR daR TR TR′ DαR DαR′ faL charge 000 0 0 1 1 1 1 1 1 1 1 0 B 3 3 3 3 3 3 3 3 charge 1 1 1 1 0 11 0 0 2 2 2 2 0 L − − − − −

TABLE II: Nonzero lepton number L of the Higgs ﬁelds in the 3-4-1 model with right-handed neutrinos.

0 ++ + + 0 0 0 + + ++ ++ Fields χ1 χ2− φ1− φ−− ρ ρ2 η3 η4 H1 H4 H1 H4 H1 H2 L 2 2 2 2 2 2 2 2 2 2 2 2 2 2 − − − − − − − − −

The above formula is useful for the extensions where the ﬂavor discrete symmetries such as

A4,S3, etc, are implemented. c c Now we return to our model, where the lepton quadruplet in (63) contains la and νa with lepton number ( 1). We then get − 4 2√6 β = ,γ = . (86) √3 3 Hence, the lepton number operator in the minimal 3-4-1 model with right-handed neutrinos gets the form 4 1 L = T + T + . (87) √ 8 √ 15 L 3 2 This formula is an extension of that in the 3-3-1 model [33]. For the quadruplet, this operator has the form L = diag (1 + , 1+ , 1+ , 1+ ) . (88) L L − L − L The ﬁelds with nonzero lepton numbers are listed in Tables I, II, and III.

TABLE III: Nonzero lepton number L of fermion in the 3-4-1 model with right-handed neutrinos.

Fields la νa T T ′ Dα Dα′ L 1 1 2 2 2 2 − −

22 Now we turn to the gauge boson sector. The contribution to the gauge boson masses from H arises from a piece,

H = Tr[(D H )+(Dµ H )] Lmass µh i h i 2 CC µCC NC µNC = g Tr[(P H )†(P H )+(P H )†(P H )], (89) µ h i h i µ h i h i

where

15 D = ∂ ig A T ig′XB′′T µ µ − aµ a − µ 16 a=1 X ∂ igP ≡ µ − µ ∂ igP NC igP CC. (90) ≡ µ − µ − µ

As a result of the symmetric form of the two quadruplets, we have (for details, see [34])

k k (PµH)ij = (Pµ)i Hkj +(Pµ)j Hki . (91)

For the gauge boson masses, one needs to calculate

(P H ) = (P )k H +(P )k H . (92) µh i ij µ i h ikj µ j h iki

We ﬁrst deal with the charged gauge boson masses being deﬁned through

1 P CC = λ A , a =1, 2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14 µ 2 a a a X ′+ 0 ′+ ′ 0 0 W W13− W14 0 W Y − N ′ ′ ′ 1  W − 0 W23−− W24−  1  W − 0 U −− X −  = = ′ ′ , (93) √2  W + W ++ 0 W +  √2  Y + U ++ 0 K +   13 23 34     0 +   0 ′+ ′   (W )∗ W W − 0   (N )∗ X K − 0   14 24 34 µ  µ     ′+ ′ ′ ′+ + where we have denoted √2W A iA , Y − W − ,X − W − ,K W , U −− µ ≡ 1µ − 2µ ≡ 13 ≡ 24 ≡ 34 ≡ 0 0 W −− and N W . 23 ≡ 14 0 The masses of the non-Hermitian neutral N and doubly charged U ±± gauge bosons are

2 2 2 2 2 2 2 2 2 g (ω + v +4v′ ) 2 g (V + u +4ǫ ) m ±± = , m 0 = , (94) U 4 N 4

0 0 where the N and N ∗ gauge bosons do not mix with other Hermitian neutral gauge bosons. The squared mass matrix of the singly charged gauge bosons is rewritten in the basis

23 ′ ′ ′ ′ T (W ±,K ±,X ±,Y ±) as follows:

2 2 2 v + u + v′′ 2v′ǫ 0 0

2 2 2 2 2 g  w + V + v′′ 0 0  MG± = , (95) 4  v2 + V 2 + v 2 2v ǫ   ′′ ′   2 2 2   u + w + v′′      2 2 2 where v′′ v′ + ǫ . In the limit ǫ = 0, all of the above masses are the same as those given ≡ in [9] (with unique diﬀerences associated with v′′, since in [9] the authors did not use the redeﬁned decuplet). Note that all nondiagonal elements of the matrix (95) are proportional

to v′ǫ, which are much smaller than the diagonal ones; therefore the mass eigenstates of the singly charged gauge bosons can be reasonably identiﬁed with those given in [9]. The physical states are determined as

W = cos θ W ′ sin θK′ , K = sin θ W ′ + cos θK′ , (96) µ µ − µ µ µ µ where the W K mixing angle θ characterizing lepton number violation is given by − 4v ǫ tan2θ = ′ . (97) V 2 + ω2 u2 v2 − − For the X Y mixing, we obtain the physical states, −

Y = cos θ′ Y ′ sin θ′ X′ , X = sin θ′ Y ′ + cos θ′ X′ (98) µ µ − µ µ µ µ

with the mixing angle θ′ deﬁned as

4v′ǫ tan2θ′ = . (99) V 2 ω2 u2 + v2 − − The masses of the physical states are determined as

2 2 2 g 2 2 2 2 g 2 2 2 m ± (v + u + v′′ ) , m ± (V + w + v′′ ) , W ≃ 4 K ≃ 4 2 2 2 g 2 2 2 2 g 2 2 2 m ± (V + v + v′′ ) , m ± (w + u + v′′ ) . (100) X ≃ 4 Y ≃ 4 From (100), it follows that

2 2 2 2 2 v + u + v′′ v = (246GeV) , (101) ≃ SM and bounds on the singly charged gauge boson mass splitting are

m2 m2 m2 , m2 m2 m2 m2 . (102) | K − X | ≤ Y | K − X − Y | ≤ W 24 Comparing (97) with (99), we see that the mixing angle between the lightest W and heaviest K is smaller than the X Y mixing angle. The mixing angle is quite small and can be − constrained from the W decay width (as in the economical 3-3-1 model [35, 36]). From the experimental point of view, the approximation in previous works [9, 17], V = ω creates the diﬃculty in distinguishing between the bileptons X and Y . So the natural way

is to assume V ω. Note that U ±± and Y ± are similar to the singly charged gauge bosons ≫ 0 in the minimal 3-3-1 model [3], while N and X± play the similar role in the 3-3-1 model

with right-handed neutrinos [2, 4]. The heaviest singly charged gauge bosons K± are the completely new ones that couple with the exotic quarks and right-handed leptons only (see, Sec. III A). In our assignment (and also in Voloshin’s paper [8]), particles belonging to the minimal version are lighter than those in the 3-3-1 model with right-handed neutrinos [i,e., Eqs.(94] and (100)). For the original 3-4-1 model [2, 9], the above consequence is the opposite.

Now we turn to the neutral gauge boson sector. In the basis A3µ, A8µ, A15µ, Bµ′′ , the squared mass matrix for the neutral gauge bosons is given by

g2 M 2NG = mass 4

2 2 2 1 2 2 2 1 2 2 2 t 2 u + v + v′′ (u v + v′′ ) (u v 2v′′ ) v √3 − √6 − − − √2   1 2 2 2 2 1 2 2 2 2 t 2 2 3 (u + v +4ω + v′′ ) √ (u + v 2ω 2v′′ ) √ (v +2ω )  3 2 − − 6  . (103) ×  1 2 2 2 2 2 t 2 2   (u + v + ω +9V +4v′′ ) (v ω )   6 2√3 −   2   t 2 2   (v + ω )   2    2NG It has a property that Det(Mmass ) = 0, implying a massless state of the photon. The mass eigenvalues of this matrix can be done the same way as in the general case. For this particular case, we have

2√2 √3t s43 = 1, c43 =0, s32 = , c32 = − , √8+3t2 √8+3t2 2√8+3t2w2 t = . (104) 2α 9V 2 (7+3t2)w2 − T T The transformative matrix C41 satisfying (A,Z,Z3′ ,Z4′ ) = C41.(A3, A8, A15, B′′) now has

25 the form

sW cW c32 0 cW s32

 cW sW c32 0 sW s32  C41 = − − . (105)  0 c s s c c   α 32 α α 32   − −   0 s s c s c   α 32 − α − α 32    The squared mass matrix in the new basis is obtained as

00 0 0 2 2 2 2NG  mZ m23 m24  M41 = , (106) 2 2  m ′ m   Z3 34   2   m ′   Z4    where

2 2 2 2 2 2 g (v + u + v′′ ) mW mZ = 2 = 2 , 4cW cW 2 2 2 2 2 g 2 2 2 2 g 2 m ′ = 9s V + sα cα 8+3t w + √2cαs32 + sα u Z3 24 α − 24 p 2 2 2 (3t + 4)cαs32 2 2 + sα + v +2 √2sα cαs32 v′′ , 2√2 − # 2 2 2 2 2 g 2 2 2 2 g 2 m ′ = 9c V + cα + sα 8+3t w + cα √2sαs32 u Z4 24 α 24 − p 2 2 2 (3t + 4)sαs32 2 2 + cα v +2 √2cα + sαs32 v′′ , − 2√2 # 2 2 2 g cα 1 4s sα cα(3 2c ) sα m2 = − W + u2 + − W + v2 23 2 2 2 4 "− cW √3 √6cW ! c 3(1 4s ) √6cW ! p W − W 2 cα 1 4sW √2sα 2 2 p + − + v′′ (m ), − 2 √ √ ∼ O W pcW 3 3cW ! # 2 2 2 2 2 2 2 2 2 g cα(v u +2v′′ ) sα (4c 3)(u + v′′ )+(2c 3)v m2 = − + W − W − (m2 ), 24 2 2 W 4 " √6cW c 3(4c 3) # ∼ O W W − 2 2 g c2α 2 2 2 2p m = 4(u 2v′′ )+(3t + 4)v 34 24 √ t2 − 3 +8 s2α 2 2 2 4 2 2 2 2 + (3t 8)u (8+21t +9t )v + 4(3t + 4)v′′ (m ). (107) 2(3t2 + 8) − − ∼ O W It can be seen that all of the nondiagonal elements are in the order of (m2 ). Therefore, O W 2 2 they are much smaller than m ′ and m ′ , implying that these two values can be approx- Z3 Z4 imately the eigeinvalues of the matrix (106). Furthermore, the largest contribution to the 2 2 2 2NG 2 u +v m = M in the ﬁnal diagonal matrix, which is proportional to m ( 2 2 ), is Z 41 22 W × O V +w also small. In conclusion, the matrix (106) can be considered as the diagonal matrix where

26 the eigenvalues correspond to the diagonal elements, and the matrix C41 is the one relating the two original and mass bases of the neutral gauge bosons.

A. Currents

From the Lagrangian of the fermion

µ Lfermion = i fγ Dµf, (108) f X one gets the interactions of the charged gauge bosons with leptons in the following La- grangian part:

g µ ′+ ′ c 0 c Lleptons = νaLγ W laL + Y −l + N ν √2 µ µ aL µ aL h ′ ′ µ c c c µ + c + laLγ Uµ−−laL + Xµ−νaL + laLγ Kµ νaL + H.c.

g µ + 0 c µ ′ i c = νaLγ W ′ laL + N ν laRγ Y −ν √2 µ µ aL − µ aR hµ c ′ c µ ′+ + l γ U −−l + X −ν ν γ K l + H.c. aL µ aL µ aL − aR µ aR g µ 0 c µ c i = νaLγ N ν + laLγ U −−l √2 µ aL µ aL + ν γµh(c P + s P ) l W + + ν γµ (s P c P ) l K+ a θ L θ R a µ a θ L − θ R a µ µ c µ c + l γ (c ′ P s ′ P ) ν X− l γ (s ′ P + c ′ P ) ν Y − + H.c., (109) a θ L − θ R a µ − a θ L θ R a µ where we have used lc γµνc = ν γµl . aL aL − aR aR

From (109), we see that new gauge bosons K± play a similar role of the SM W ± for right- handed leptons with just opposite sign of coupling constant g. This right-handed current also appeared in [37]. The gauge bosons carrying lepton number 2 (called bilepton gauge 0 bosons) include: Y ±, N , U ±±, and X±. For quarks, we have

g µ ′+ ′ 0 Lquarks = u3Lγ W d3L + Y −TL + N T ′ √2 µ µ µ L hµ ′ µ ′+ + d3Lγ Uµ−−TL + Xµ−TL′ + TLγ Kµ TL′

µ ′ ′+ 0 + d γ W −u + Y D + N ∗D′ αL − µ αL µ αL µ αL µ ++ ′+ µ ′ u γ U D + X D′ + D γ K −D′ + H.c.. (110) − αL µ αL µ αL αL µ αL i 27 Taking into account the mixing among singly charged gauge bosons, we can express the above expression as follows,

CC g µ + µ + µ + µ + µ0 0 µ ++ = J −W + J −K + J −X + J −Y + J ∗N + J −−U + H.c. (111), −L √2 W µ K µ X µ Y µ N µ U µ where

µ µ µ µ J − = c (ν γ l + u γ d u γ d ) W θ aL aL 3L 3L − αL αL µ µ µ s ( ν γ l + T γ T ′ + D′ γ D ), (112) − θ − aR aR L L αL αL µ µ µ µ µ µ µ J − = c ( ν γ l + T γ T ′ + D γ D )+ s (ν γ l + u γ d u γ d ), K θ − aR aR L L αL′ αL θ aL aL 3L 3L − αL αL µ c µ µ µ c µ µ µ J − = c ′ (ν γ l + T γ d u γ D′ )+ s ′ (l γ ν + T γ u + d γ D ) , X θ aL aL L′ 3L − αL αL θ aL aL L 3L αL αL µ c µ µ µ c µ µ µ J − = c ′ (l γ ν + T γ u + d γ D ) s ′ (ν γ l + T γ d u γ D′ ) , Y θ aL aL L 3L αL αL − θ aL aL L′ 3L − αL αL µ c µ µ µ J −− = l γ l + T γ d u γ D , U aL aL L 3L − αL αL µ0 µ c µ µ JN ∗ = νaLγ νaL + u3Lγ TL′ + DαL′ γ dαL. (113)

It is emphasized that in the above expression, all fermions are in the weak states. For precision, they should be in the mass states. For the latter case, in the quark sector, the CKM matrix will appear. In the model under consideration, due to the neutrino Dirac µ mass matrix, the lepton mixing matrix VPMNS will appear in the JW−. So in terms of mass eigenstates, the current in (112) has a new form:

µ µ ij µ ij J − = c (ν γ V l + s ν γ V l )+ . (114) W θ iL PMNS jL θ iR PMNS jR ···

The neutral currents, including the electromagnetic current, are

3 NC µ g4 i ¯ µ (V ) (A) = eJemAµ + Zµ fγ [g (f)iV g (f)iAγ5]f , (115) −L 2cW { − } i=1 f X X where g′ 2√2 sin θW e = g sin θW , t = = , (116) g 1 4 sin2 θ − W 1,2,3 1 2,3 and Z , which can be identiﬁed as Z Z andpZ Z′ , , are exact eigenstates of the ≃ ≃ 3 4 matrix (106). The neutral currents are similar to that shown in Ref. [9], so the reader is referred to the mentioned work. Similar to the 3-3-1 models [38], in the model under consideration, there are FCNCs at the tree level due to Z2 and Z3.

28 The formula (116) leads to a consequence,

2 sin θW < 0.25, (117) which is the same as in the minimal 3-3-1 model. As mentioned in (56), this constraint is the same for the original version [2, 9] and for the version we are considering now. With the particle content in both the fermion and Higgs sectors similar to that in [23], we suggest that the uniﬁcation mass is in the range of (10) TeV. The possible Landau pole O similar to those in the minimal 3-3-1 model [39] will be considered in our future work. We note that some interesting aspects relating to the Landau poles of both the minimal 3-3-1 models and 3-4-1 models were indicated in [19].

IV. HIGGS POTENTIAL

The most general potential can then be written in the following form:

V (η,ρ,φ,χ,H)= V (η,ρ,φ,χ)+ V (H), where

2 2 2 2 V (η,ρ,φ,χ) = µ1η†η + µ2ρ†ρ + µ3φ†φ + µ4χ†χ

2 2 2 2 + λ1(η†η) + λ2(ρ†ρ) + λ3(φ†φ) + λ4(χ†χ)

+ (η†η)[λ5(ρ†ρ)+ λ6(φ†φ)+ λ7(χ†χ)]

+ (ρ†ρ)[λ8(φ†φ)+ λ9(χ†χ)] + λ9′ (φ†φ)(χ†χ)

+ λ10(ρ†η)(η†ρ)+ λ11(ρ†φ)(φ†ρ)+ λ12(ρ†χ)(χ†ρ)

+ λ13(φ†η)(η†φ)+ λ14 (χ†η)(η†χ)+ λ15 (χ†φ)(φ†χ)

ijkl + (fǫ ηiρjφkχl + H.c.), (118) and V (H) consists of the lepton-number-conserving (LNC) and -violating (LNV) parts, namely,

V (H) V + V , (119) ≡ LNC LNV 2 2 2 VLNC = µ5Tr(H†H) + [f4χ†Hη∗ + H.c.]+ λ16Tr[(H†H) ]+ λ17[Tr(H†H)]

+Tr(H†H)[λ18(η†η)+ λ19(ρ†ρ)+ λ20(φ†φ)+ λ21(χ†χ)]

+λ22(χ†H)(H†χ)+ λ23(η†H)(H†η)+ λ24(ρ†H)(H†ρ)+ λ25(φ†H)(H†φ),

VLNV = f2χ†Hχ∗ + f3η†Hη∗ + λ26Tr(H†H)(χ†η)+ λ27(χ†H)(H†η) + H.c.. (120)

29 In the below illustration for the Higgs spectrum we consider only the LNC part of V (H). The minimum conditions correspond to the six linear coeﬃcients of the neutral Higgs bosons with non-zero VEVs that vanish, leading to the six equalities shown in Appendix A. The squared mass matrix 2 of the doubly charged Higgs (DCH) bosons is shown in MDCH Appendix A. It can be checked that det 2 =0, so that there exist two Goldstone bosons MDCH of the U ±± bosons. They can be found exactly as follows:

√2v′H1±± √2v′H2±± vρ±± + wφ±± G±± = − − . (121) U 2 2 2 √w + v +4v′

2 There are three physical masses. In the limit v′ 0, these masses are ≃ 2 2 2 1 2 2 2 2 w + v fVu m ±± = ( λ25w + λ24v )= m ±± , m ±± = λ11 , (122) h1 4 − − h2 h3 2 − wv where hi±±, i = 1, 2, 3 are mass eigenstates of the DCHs. Hence, in the limit v′ = 0, there always exists a negative value of 1 ( λ w2 + λ v2) , implying a negative squared mass at −| 4 − 25 24 | 2 the tree level. On the other hand, when v′ = 0, the matrix in (A2) gives a tree-level 6 MDCH 2 3 2 ±± mass relation, Tr( DCH)= i=1 mh , or equivalently, M i

P 2 2 2 2 2 2 w + v fVu m ±± + m ±± + m ±± = λ16v′ + λ11 . (123) h1 h2 h3 2 − wv As a consequence of the vacuum stabilities that the Higgs potential must be bounded from

below, we have λ16 > 0. Then the sum of the two squared DCH masses (123) is in the order 2 2 2 of (λ v′ ). Because the DCH are solutions of the equation det I m ±± = 0, O 16 MDCH − 4 × h we have another relation:

2 2 2 2 2 2 1 2 2 2 w + v +4v′ fVu m ±± m ±± m ±± = ( λ25w + λ24v ) λ11 . (124) h1 h2 h3 −16 − × 2 − wv The right-hand side of (124) is nonpositive because the factor λ fVu has the same 11 − wv positive sign with squared masses of the heavy DCH h3±±. Hence, there is always a negative squared mass of DCH at the tree level. To avoid DCH tachyons, the λ w2 λ v2 should | 25 − 24 | be small so that the loop contributions can raise the DCH mass to positive values. As a result, the parameter λ should be very small, and the model predicts the existence of | 25| rather light DCHs. There are 12 pairs of singly charged Higgs (SCH) components in the original basis. In the mass basis, there are four massless pairs which are Goldstone bosons of W, X, Y, and K

30 gauge bosons. In the limit ǫ = 0, the squared mass matrix in the original basis decomposes into four independent 3 3 matrices; see the details in the Appendix. Each of them has × only one zero eigenvalue, implying the massless state, and two other massive values. All of the massless states are

v′H1± wφ1± + uη3± V χ2± + v′H4± + vρ2± G± = − − , G± = − , 1 √ 2 2 2 2 √ 2 2 2 w + u + v′ V + v + v′ wφ2± V χ3± + v′H3± uη2± + vρ1± + v′H2± G± = − , G± = − . (125) 3 √ 2 2 2 4 √ 2 2 2 V + w + v′ u + v + v′

In this limit, it is easily to identify that G± G±, G± G±, G± G±, and G± G± Y ≡ 1 X ≡ 2 K ≡ 3 W ≡ 4 which are respective Goldstone bosons absorbed by Y ±, X±, K±, and W ± bosons.

With v′ = 0, the masses as well as mass eigenstates are a bit complicated. For illustration, 6 it is enough to consider here the limit v′, ǫ 0. The mass eigenvalues of the other eight → pairs of SCHs are

2 1 2 2 2 1 2 2 m ± = (λ23u λ25w ), m ± = λ23u λ24v , h1 4 − h2 4 − 2 2 2 2 2 u + v wV 2 u + ω fvV m ± = (λ10 f ), m ± = λ13 , h3 2 − uv h4 2 − wu 2 2 2 2 2 V + ω fvu 2 V + v fwu m ± = λ15 , m ± = λ12 , h5 2 − Vw h6 2 − V v 2 1 2 2 2 1 2 2 m ± = λ22V λ25w , m ± = λ22V λ24v (126) h7 4 − h8 4 − with the respective mass states as follows:

vη2± + uρ1± uφ1± + wη3± h± H±, h± = H±, h± = , h± = , 1 ≡ 1 2 2 3 √u2 + v2 4 √u2 + w2

Vφ2± + wχ3± vχ2± + V ρ2± h± = , h± = , h± H±, h± H±. 5 √V 2 + w2 6 √V 2 + v2 7 ≡ 3 8 ≡ 4

The model predicts two rather light SCHs, h1± and h2±, because λij should be in the order

of (1); λ w is not too large; and u, v′ and v are in the electroweak scale. O | 25 | There are ten neutral Higgs components in the original basis. Four of the ten CP-odd neutral Higgses are massless, of which the four independent combinations of them are four Goldstone bosons of the Z, Z2 Z3 and N 0 gauge bosons. But there is still one more massless state, which is exactly H0, at the tree level. In the limit ǫ 0, the mass eigenstates of the 3 → CP-odd neutral Higgs are shown explicitly in Appendix A. There are ﬁve massive states, with 0 eigenstates denoted as HAi (i =1, 5), where three imagined parts of H1,2,4 are approximate

31 egeinstates. The condition of the positive mass of the CP-odd neutral Higgs HA5 shows that f < 0. There also exists one light CP-odd neutral Higgs boson. In the neutral sector, one of the ten CP-even neutral Higgs bosons is the Gold- 0 stone boson of the N ∗ boson. The squared mass matrix separates into two subma- 2 2 trices, namely the 4 4 and 6 6 matrices. They are denoted as 0 and 0 , × × M1H M2H 0 0 0 0 T corresponding to the two respective sub-bases (Re[H1 ], Re[H4 ], Re[χ1], Re[η4]) and 0 0 0 0 0 0 T (Re[H3 ], Re[χ4], Re[φ3], Re[ρ2], Re[η1], Re[H2 ]) . The massless values are contained 2 in 0 . In the limit ǫ 0, three other mass values are M1H → 2 2 2 1 2 2 2 2 2 V + u fwv m 0 = 2λ23u λ24v 2λ16v′ λ25w , m 0 = λ14 , h1 4 − − − h2 2 − Vu 2 1 2 2 2 2 m 0 = 2λ22V λ24v 2λ16v′ λ25w . (127) h3 4 − − − 0 0 0 0∗ The mass eigenstates (h1, h3, h3) and the Goldstone bosons GN in this case are

V u 0 G 0∗ Re[χ ] 0 0 0 0 N √V 2+u2 √V 2+u2 1 h1 Re[H1 ], h3 Re[H4 ], = − . ≡ ≡  h0   u V   Re[η0]  2 √V 2+u2 √V 2+u2 4       It can be seen that while the two last are very heavy with the order of the SU(3)L and SU(4) breaking scales, the ﬁrst Higgs boson may be lighter because λ w2 should not be L | 25 | large as discussed above. So it may be the SM Higgs boson or that in the 750 GeV diphoton excess. 2 Now consider the second mass matrix 0 . From our investigation, in general it is M2H 2 2 easy to check that det 0 = 0. But if v′ = 0, 0 has a massless value. In addition, M2H 6 M2H if v′ = v = u = 0, the matrix has two massless values, implying that there may be two light CP-even neutral Higgs bosons. Hence one of them can be identiﬁed with the SM Higgs boson, meaning that the Higgs sector of the model under consideration is reliable. The main contributions to the four heavy Higgs bosons are

2 2 1 2 2 m 0 = fwV, m 0 = λ22V λ25w , h4 − h5 4 − 2 2 2 2 2 2 2 2 2 m 0 = λ3w + λ4V ( λ3w + λ4V ) + λ9′ w V . (128) h6,7 ± − q To conclude, in the Higgs sector, we would like to emphasize two important results. First, the above investigation can be applied for the models where the 10S H is not included. Second, the model predicts many Higgs bosons with masses near the TeV range that today colliders can detect. Hence this aspect of the Higgs sector should be explored in more detail.

32 V. PHENOMENOLOGY

Our aim in this section is to ﬁnd some constraints on the parameters of the model. From mixing of the singly charged gauge bosons, we have some special features related to the SM W boson.

1. In the model under sonsideration, the W boson has the following normal main decay modes:

W − l ν˜ (l = e, µ, τ), → l ucd,ucs,ucb, (u c), (129) ց → which are the same as in the SM. Due to the W K mixing, we have other modes − related to right-handed lepton counterparts, namely,

W − l ν˜ (l = e, µ, τ). (130) → R lR It is easy to compute the tree-level decay widths as follows [40]. The predicted total width for the W decay into fermions is

tot αMW 2 αMW ΓW =1.04 2 (1 sθ)+ 2 . (131) 2sW − 4sW This is quite similar to the case of the economical 3-3-1 model [36]. From the recent tot data of the W ± boson [32]: α(m ) 1/128, m = 80.385 0.015 GeV, Γ = Z ≃ W ± W 2.085 0.042 GeV. The s is less constrained than that of [36], with an upper bound ± θ of s 0.19. θ ≤ 2. In the model under consideration, the muon decay,

µ− e− +˜ν + ν → e µ , consists of two diagrams mediated by W and K. The Feynman diagrams are on the left in Fig. 1.

The decay width is given by 4 5 g mµ 1 1 Γ(µ− e− +˜ν + ν )= + . (132) → e µ 6144π2 × m4 m4 W K Because the model predicts the wrong decay µ− e− + ν +˜ν , we assume that the → e µ total total decay of the muon is Γ = Γ(µ− e− +ν ˜ + ν )+Γ(µ− e− +ν ˜ + ν ). µ → e µ → µ e This result will be used for in a future study.

33 e− e−

µ− W, K µ− X,Y

ν νc ˜e c e νµ νµ

FIG. 1: Feynman diagram giving contribution to muon decay. The left and right diagrams present the main and wrong decay channels, respectively.

3. The wrong decay of the muon is

µ− e− + ν +˜ν , → e µ

where the Feynman diagram is on the right side of Fig. 1. The branching ratio of the wrong decay Br(µ e ν ν¯ ) < 0.012 leads to the following constraint: → e µ 1 1 4 + 4 mX mY 1 1 1 1 1 0.012 1 1 1 1 < 0.012 or 4 + 4 0.012 4 + 4 + 4 < 4 . m4 + m4 + m4 + m4 mX mY − mK mX mY mW W K X Y (133) In the limit V w, i.e., 1/m4 , 1/m4 1/m4 , we obtain the below constraint of ≫ K X ≪ Y m : m > m √4 82.333 242 GeV. This is consistent with those in Ref. [41]. In Y Y W × ≃ the limit of V w, implying that m2 m2 m2 /2, the constraint is more strict: ≃ X ≃ Y ≃ K m > m √4 164.417 287 GeV. Y W × ≃ 4. The µ e conversion: The charged current of the model under consideration has a − similar structure as the ones discussed in [41–44] so these have the similar results as

the µ e conversion. In particular, the mass of the doubly charged bileptons U ±± − satisﬁes m ±± 135 GeV [43]. Note that this result was concluded for the SU(3) , U ≥ L but it is the same for the SU(4)L because they have the same gauge couplings. This

constraint is less strict than the constraint from the wrong muon decay, because mY ±

and mU ±± are related with only the SU(3)L breaking scale w.

34 VI. CONCLUSION

In this paper, we have analyzed the 3-4-1 model with arbitrary electric charges of the extra leptons. The scalar and gauge boson sectors are presented in detail . The mixing matrix, the eigenmasses, and the eigenstates of neutral gauge bosons are analyzed. For future studies, we will focus on the scalar sector, especially neutral one, in which the SM-like Higgs boson is contained. Next, we have presented a new development of the original 3-4-1 model. Diﬀerent from previous works [2, 9], in this paper, with the assumption of a new nonzero VEV of a neutral Higgs component in the decuplet H, some new interesting features occur: (i) the neutrinos get Dirac masses at tree level; ii) the mixing among the singly charged gauge bosons leads to the appearance of a new small contribution to the decay width of the W boson, the same as that shown in the economical 3-3-1 model. But under the present W data, the constraint of the mixing angle is less strict. The model also predicts the existence of many bileptons, including new quarks, and gauge and Higgs bosons, as well as the LFV interactions. Like the 3-3-1 models, many of these bileptons contribute to the LFV processes such as the wrong muon decay, and the µ e conversion. If the SU(4) breaking scale is much larger than the − L SU(3)L breaking scale, so that it gives suppressed contributions to the LFV processes, the constraint of the SU(3)L breaking is the same as what was indicated in the 3-3-1 models.

In contrast, if the two breaking scales are close together, the lower bounds of the SU(3)L breaking scale signiﬁcantly increase. We see this point in the case of the wrong muon decay, where the lower constraints of mY are 242 and 287 GeV, corresponding to the two mentioned cases. We have also derived the lepton number operator, and the lepton number of the ﬁelds in the model is presented. As in the minimal 3-3-1 model, in the 3-4-1 model with right-handed neutrino considered 2 here, there exists a bound on the sine-squared of the Weinberg angle, namely, sin θW < 0.25. The constraint on the electric charges of extra leptons has been obtained as well. The Higgs sector is roughly studied. In the limit of lepton number conservation, the Higgs sector contains all massless Goldstone bosons for massive gauge bosons and the SM- like Higgs boson. The model we have considered is quite interesting and deserves further study.

35 Acknowledgments

LTH thanks Le Duc Ninh for useful discussions on gauge anomalies and SM matching conditions. This research is funded by the Vietnam National Foundation for Science and Technology Development (NAFOSTED) under grant number 103.01-2014.51.

Appendix A: Higgs spectrum

This section pays attention to the Higgs potential satisfying the lepton number conservation. In addition, the squared mass matrices are written in the limit of ǫ = 0, except the squared mass matrix of the doubly charged Higgs which is independent of ǫ.

1. Minimal conditions of the Higgs potential

We list here six equalities for minimal conditions of the Higgs potential:

2 1 2 2 2 2 2 2 2 2 2 µ = 2λ ǫ (v′ ǫ ) λ ǫ V + λ ǫ w + λ ǫ v +2fVwvu 1 −4 16 − − 22 25 24 1 2 2 2 2 2 λ v′′ +2λ u + λ v + λ w + λ V , − 2 18 1 5 6 7 2 1 2 2 2 2 fwVu 2 1 2 µ = 2λ v + λ u + λ w + λ V + + λ v′′ + λ v′ , 2 −2 2 5 8 9 v 19 2 24 2 1 2 2 2 2 fuVv 2 1 2 µ = 2λ w + λ u + λ v + λ′ V + + λ v′′ + λ v′ , 3 −2 3 6 8 9 w 20 2 25 2 1 2 2 2 2 2 µ = 2λ V + λ u + λ v + λ′ w + λ v′′ 4 −2 4 7 9 9 21 1 2 2 2 2 2 2 2 2 2 2fVwuv +2λ ǫ (v′ ǫ ) λ ǫ u + λ ǫ v + λ ǫ w , − 4V 2 16 − − 23 24 25 2 1 2 2 2 1 2 2 2 1 2 µ = λ v′ +2λ v′′ + λ u +(λ + λ )v + λ w + λ V + λ w , 5 −2 16 17 18 19 2 24 20 21 2 25 ǫ 2 2 2 2 2 2 f = 2λ (v′ ǫ ) λ V λ u + λ v + λ w . (A1) 4 2Vu 16 − − 22 − 23 24 25

36 2. Squared mass matrix of doubly charged Higgses

T Squared mass matrix of the doubly charged Higgses in the basis (H1±±, H2±±, ρ±±, φ±±) is given by

2 2 2 2 2λ v′ + λ v λ w 2λ v′ √2λ vv′ √2λ wv′ 16 24 − 25 16 24 25   2 2 2 1 2λ16v′ λ24v + λ25w √2λ24vv′ √2λ25wv′ 2 =  −  . MDCH 4    2λ w2 2fwVu 2λ wv 2fVu   11 v 11   − −   2 2fVvu   2λ11v   − w   (A2)

3. Squared mass matrices of singly charged Higgses

The squared mass matrix of the singly charged Higgs consists of two independent 6 6 matrices. They are denoted as 2 and 2 with respective sub-bases × M1sch M2sch T T (H1±, φ1±, η3±, χ2±, H4±, ρ2±) and (φ2±, χ3±, H3±, η2±, ρ1±, H2±) . In the limit ǫ = 0, they divide into four independent 3 3 submatrices, denoted as × 2 2 2 1sch′ 0 2 2sch′ 0 1sch = M , and 2sch = M , 2 2 M  0 ′′  M  0 ′′  M1sch M2sch     where

2 2 λ u λ w λ wv′ λ uv′ 23 − 25 25 23 1   2 2 2 2fV vu 1sch′ = 2λ13u λ25v′ 2(λ13wu fV v) , (A3) M 4  − − w −     λ w2 λ v 2 2fVwv   2 13 + 23 ′ u   −   

2 2 2fwuv λ v′ +2λ v λ V v′ 2 (λ V v fwu) 22 12 − V 22 12 − 1   2 2 2 ′′1sch = λ22V λ24v λ24vv′ , (A4) M 4  −     λ V 2 λ v 2 2fVwu   2 12 24 ′ v   − −   

2 2 2fVvu 2λ V λ v′ 2 (λ V w fvu) λ v′w 15 − 25 − w 15 − 25 1   2 2 2 2fwvu 2sch′ = 2λ15w + λ22v′ λ22V v′ , (A5) M 4  − V     2 2   λ22V λ25w   −    37 and

2 2 2fVwv λ v′ +2λ v 2 (λ uv fV w) λ v′u 23 10 − u 10 − 23 1   2 2 2 2fV uw 2sch′′ = 2λ10u λ24v′ λ24vv′ . (A6) M 4  − − v     2 2   λ23u λ24v   −   

4. Squared mass matrices of CP-odd neutral Higgses

This 10 10 matrix has a massless state of Im[H0], even ǫ = 0. Furthermore, the × 3 6 remaining part separates into two 4 4 and 5 5 matrices, corresponding to the bases of × × 0 0 0 0 T 0 0 0 0 0 T (Im[χ1], Im[η4], Im[H4 ], Im[H1 ]) and (Im[ρ2], Im[φ3], Im[χ4], Im[η1], Im[H2 ]) . In the limit ǫ = 0, the following Higgses are identiﬁed with mass eigenstates:

0 2 1 2 2 2 2 Im[H ] H , m = 2λ V 2λ v′ λ v λ w , 4 ≡ A1 A1 4 22 − 16 − 24 − 25 0 2 1 2 2 2 2 Im[H ] H , m = 2λ u 2λ v′ λ v λ w , 1 ≡ A2 A2 4 23 − 16 − 24 − 25 0 2 1 2 2 2 2 2 Im[H ] H , m = λ V + λ u 2λ v′ λ v λ w . (A7) 2 ≡ A3 A3 4 22 23 − 16 − 24 − 25 The nontrivial parts of the two matrices are now 2 2 and 4 4 which relate to the four × × 0 Goldstone bosons, including GN and GZi (i =1, 2, 3), and the two massive CP-odd neutral Higgses, in particular

G V u Im[χ0] N0 √V 2+u2 √V 2+u2 1   =     (A8) u V 0 HA4 2 2 2 2 Im[η4]    − √V +u √V +u    and      

v v 0 G 2 2 0 0 2 2 Im[ρ ] Z1 − √v +u √v +u 2    2 √(v2+u2V 2    vu 0 uv 0 GZ2 2 2 √A 2 2 Im[φ3]   =  −√A(v +u ) −√A(v +u )    , (A9)    V 2u2v √Aw V v2u2 V 2v2u   0   GZ3     Im[χ ]     √AB √B √AB √AB   4     − − −       V wu V vu wvu V wv   0   HA5     Im[η1 ]     √B √B √B √B          where A = V 2v2 + u2(V 2 + v2), B = V 2v2(w2 + u2) + w2u2(V 2 + v2). Note that the Goldstone bosons of the three Hermitian gauge bosons are linear combinations of the three above massless states GZi , not exactly themselves. But the GZ1 mainly contributes to the

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